Q24 / 180 · KCET 2023 · Chemistry

At $500 \mathrm{~K}$, for a reversible reaction $A_2(g)+B_2(g) \rightleftharpoons 2 A B(g)$ in a closed container, $K_C=2 \times 10^{-5}$. In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant $K_C$ in the presence of catalyst at the same temperature is

  • A
    $2 \times 10^{-4}$
  • B
    $2 \times 10^{-6}$
  • C
    $2 \times 10^{-10}$
  • Correct answer
    D
    $2 \times 10^{-5}$

Solution

The value of the $K_C$ remains same as catalyst does not affects the $K_C$. Only temperature can alter the value of the $K_C$ of a given reaction. Thus, the equilibrium constant $K_C$ in the presence of catalyst at the same temperature is $2 \times 10^{-5}$.

Hence, the correct answer is option: 'D'.

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